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Obtain the zeroes of the polynomial $7x^2 + 18x - 9$. Hence, write a polynomial each of whose zeroes is twice the zeroes of given polynomial.
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$7x^2 + 18x - 9 = (7x - 3)(x + 3)$
$\therefore$ Zeroes are $-3, \frac{3}{7}$
New zeroes are $-6, \frac{6}{7}$
Sum of new zeroes $= (-6) + \frac{6}{7} = -\frac{36}{7}$
Product of new zeroes $= (-6) \times \frac{6}{7} = -\frac{36}{7}$
$\therefore$ Required polynomial is $x^2 + \frac{36}{7}x - \frac{36}{7}$ or $7x^2 + 36x - 36$
$\therefore$ Zeroes are $-3, \frac{3}{7}$
New zeroes are $-6, \frac{6}{7}$
Sum of new zeroes $= (-6) + \frac{6}{7} = -\frac{36}{7}$
Product of new zeroes $= (-6) \times \frac{6}{7} = -\frac{36}{7}$
$\therefore$ Required polynomial is $x^2 + \frac{36}{7}x - \frac{36}{7}$ or $7x^2 + 36x - 36$